Question

A solution of is electrolytically reduced to Mn3+. A current of 5.19 A is passed through the solution for 15.0 min. What is the number of moles of Mn3+ produced in this process? (1 faraday = 96,486 coulombs)

a.
0.145 mol
b.
0.0161 mol
c.
2.69 × 10-4 mol
d.
0.0484 mol
e.
1.08 × 10-3 mol

Answer 1

Explanation :

To find the number of moles of Mn3+ produced, we need to calculate the amount of charge passed through the solution.

The charge (in coulombs) can be calculated using the formula:

Charge = Current × Time

Converting the time from minutes to seconds:

Time in seconds (s) = 15.0 min × 60 s/min = 900 s

Charge = 5.19 A × 900 s = 4671 coulombs

Now, we can use Faraday's constant to convert the charge to moles of Mn3+.

1 Faraday (F) = 96,486 coulombs

Moles of Mn3+ = Charge / (Faraday × 3) [since 3 moles of electrons are required to reduce one mole of Mn7+ to Mn3+]

Moles of Mn3+ = 4671 coulombs / (96,486 coulombs/F) × 3

Moles of Mn3+ ≈ 0.0161 mol

Therefore, the number of moles of Mn3+ produced in this process is approximately 0.0161 mol.

The correct answer is b) 0.0161 mol.

Answer 2

Okay, let's solve this step-by-step:

* A current of 5.19 A is passed for 15.0 min = (5.19 A) * (15.0 min * 60 s/min) = 5190 sec

* The charge passed is current multiplied by time = (5.19 A) * (5190 s) = 26920 C

* 1 Faraday = 96,485 C

* Moles of electrons = charge passed/Faraday's constant

= 26920 C / 96485 C/mol

= 0.279 mol electrons

* Since 1 mol of Mn3+ requires 3 moles of electrons, the moles of Mn3+ produced will be:

0.279 mol electrons/3 electrons per Mn3+ = 0.0931 mol Mn3+

The answer is:

e. 1.08 × 10-3 mol